Question

Without using logarithm tables, evaluate: ${\log }_2{\log }_2{\log }_{2}16$

Answer 1

${\log }_2{\log }_2{\log }_{2}16={\log }_2{\log }_2{\log }_2{2}^4$

$={\log }_2{\log }_{2}4{\log }_{2}2$ .... $[\because \log a^b=b\log a]$

$={\log }_2{\log }_{2}4$ ..... $[\because {\log }_{a}a=1]$

$={\log }_2{\log }_2{2}^2$

$={\log }_{2}2{\log }_{2}2$

$={\log }_{2}2=1$